Which voltage inverting chip?

[JK Sleepling]

I'm currently using the LT1054 to invert my 9V to -9V for a bipolar supply, and while it works, I'm seeing issues. With 20 op-amps, 8 VCAs, and -5V voltage regulators in my circuit, the negative voltage drops to around -7V, likely due to the high current draw.



How do you design a stable and regulated bipolar supply?

Is there a newer, better, smaller, low-noise chip available that I might have overlooked?

[fryingpan]

The idea is (I guess, I've never used such a chip) that you power it from, say, 9V, you regulate 9V (2V drop -> +7V) and you provide a symmetrical regulated negative supply (so -7V). The stated figure is at a 500 ohm load, so 140mA.

In fact, the rest of the datasheet suggests that at 100mA the voltage loss (unregulated) is already 1.1V. So I suppose that you should consider 140mA as its top output current, aim at 100mA and settle with 1.1V voltage loss unregulated (or just go with -7V regulated). Also, you should probably use some good filtering of its output (as you have done, arguably, but the corner frequency should be about 100Hz, about 7.3 octaves away from 15kHz, about -42+dB attenuation then, so you might consider going larger), since the switching speed can get as low as 15kHz (and 25kHz can easily cause audible intermodulation issues too).

EDIT: a few corrections.

I might also note that, as regards my last point, no indication of noise or ripple is provided. That usually means "brace for impact"

[JK Sleepling]

In fact, the rest of the datasheet suggests that at 100mA the voltage loss (unregulated) is already 1.1V. So I suppose that you should consider 140mA as its top output current, aim at 100mA and settle with 1.1V voltage loss unregulated (or just go with -7V regulated). Also, you should probably use some good filtering of its output (as you have done, arguably, but the corner frequency should be about 100Hz, about 7.3 octaves away from 15kHz, about -42+dB attenuation then, so you might consider going larger), since the switching speed can get as low as 15kHz (and 25kHz can easily cause audible intermodulation issues too).

So you're saying that the LT1054 is not going to give me -9V from a 9V input, when my current draw is as high as it is? And furthermore it's noisy too?

[fryingpan]

I don't know whether the LT1054 is particularly noisy. The fact that they don't provide figures means that they are nothing to boast about, probably? And certainly the LT1054 is not going to give you -9V from a 9V input, since at 10mA output current the voltage loss (unregulated, again) is already 0.35V typical (and up to 0.55V). The regulated voltage (which I'd recommend sticking at) will reliably be -7V (within 50mV, worst case). +/-7V is not a particularly tight supply, basically all opamps you might want to use are comfortably operational at this power supply and you can count on 10 good volts for output voltage swing. (Opamps may clip at Vsupply +/- 2V).

[imJonWain]

Are you fixed on using 9VDC input? 9VAC from a wallwart could easily do what you need and more via an ac multiplier and regulator setup.

12VDC could give you +/-12V at 2W via one of those SIP package DC-DC bricks like this...
https://www.cui.com/product/dc-dc-converters/isolated/drm2-s-series

[antonis]

I'm not sure if you can use more ICs in parallel, like 7660s..



Can't recall where or when but I'm sure I've read about blown-up LT1054s in such a configuration..

There are ways to double (or triple) output current but not without significant voltage drop..
(e.g. 12V to -5V@200mA)

[merlinb]

Are you fixed on using 9VDC input?

+1. Those little switched-capacitor chips are seldom capable of much current. A circuit of this size is asking for a more appropriate power supply. Either an AC supply with biolar rectifiers, or a proper inductor-based switching regulator.

[amptramp]

merlinb is correct.  Switched capacitor filters generate a voltage spike when the capacitor is switched on or off because there is a step function in current when the input to the capacitor is suddenly switched.  With an inductor, there are no current steps during turn on because current through an inductor ramps up when voltage is applied, it doesn't come up suddenly.  There is a sudden decrease in current when the inductor is switched off but that can be handled a lot more easily with filtering.

[jorg777]

Good tips here: https://www.diystompboxes.com/smfforum/index.php?topic=132295.0

[JK Sleepling]

I really need the headroom of minimum +-8V. And still I like using a standard DC input as this is what is available in most pedal power supplies. Maybe that's a bad idea?
I can't really figure out how to make at stable bipolar 9V from Max1680. Is that possible?
Or else what chips for an inductor based circuit would you suggest in my case?

[antonis]

I can't really figure out how to make at stable bipolar 9V from Max1680. Is that possible?

Number of parallel ICs according to total current consumption and affordable voltage drop
(e.g. for 200mV drop at 50mA working current and 250mA total current draw, you need 5 devices in parallel..)

https://www.analog.com/media/en/technical-documentation/data-sheets/MAX1680-MAX1681.pdf 

P.S.
If you really want to trim +/- supplies, place a series resistor on positive supply (before voltage converters) of value -V_drop/I_total

[JK Sleepling]

Number of parallel ICs according to total current consumption and affordable voltage drop
(e.g. for 200mV drop at 50mA working current and 250mA total current draw, you need 5 devices in parallel..)

But reading the datasheet I see that max input Voltage is 6V.
How do I get +-9V  - maybe I dont understand correctly

[antonis]

My bad.. 

You could use a pair of voltage converters (voltage doubler and inverter) but it should be ridiculous..

[ElectricDruid]

Maybe that's a bad idea?

Yes, it is, but that's not your fault. It would have been nice and made life easier if pedals were powered off +/-15V (or even +/-12V) but they're not.

Many pedal supplies do have an 18V output and there are quite a few 18V pedals out there now. Perhaps you could use that? Then getting +/-9V is a simple question of creating a virtual ground / Vref like we usually do.

[fryingpan]

I'd say that with that big lot of devices you could warrant a separate power supply.

[merlinb]

Or else what chips for an inductor based circuit would you suggest in my case?

How much current do you need? I don't think you've told us. Look up DD1912PA or LT1945. It should be possible to buy a ready-made PCB that does what you need. Something like this:
https://www.ebay.co.uk/itm/295662093935?chn=ps&_ul=GB&google_free_listing_action=view_item
https://www.aliexpress.com/item/1005006480017428.html

I recommend that you also add linear regulators to the outputs, to get the voltage good and clean.

[JK Sleepling]

I need around 120 mA on the negative supply.
I changed the in/out caps from electrolytic to tantulum. Gave half a volt.

I consider regulating the +9v to 7V to hit +7V/-7V. Maybe it will do. It will still let me boost a 500mV guitar signal up to 26dB
(10V swing/ 0.5V) = gain 20

Alternatively I could maybe feed it 12V and regulate both to +9V/-9V

[R.G.]

As a step sideways...     there are some remarkably useful power controller chips these days. Here's one:
https://www.ti.com/lit/ds/symlink/ucc25800-q1.pdf?ts=1731189472344&ref_url=https%253A%252F%252Fwww.mouser.com%252F
Chip, contains all the control circuits and power devices to make a switching power supply using an off-the-shelf switching transformer; Mouser should be able to supply all the parts.

Another thing worth considering is a Recom isolated DC-DC converter. Single part, 3/8" by 1/2", $9. PCB mount in either through hole or SMD, single output of 9V unregulated or 12V unregulated, which could be then linear reg'd to whatever you like, as well as reducing any switching noise. The switching noise should be about non-existent if your layout has careful input and output decoupling caps.
https://www.mouser.com/datasheet/2/468/RN-1710937.pdf

Capacitor inverters are really intended to get small currents. You can push the designs, as noted; but it gets expensive in terms of chips, noise reduction, PCB space, design misery, and so on. I've used the Recom supplies before, and they're very handy.

If you can somehow manage to get 12V nominal into your circuit, a Mean Well N7809-2 will make +/- 9V at up to 800+ ma each. They cost $6.50 each.
https://www.mouser.com/datasheet/2/260/MeanWell_08192024_N78_2_spec-3483609.pdf

Capacitor bucket inverters are nice, but they're not the only way, especially if you're up into a watt or two of power.